Binary Stars & Binary Evolution

6
to the supernova remnant star, it is reasonable to assume that V = V o , i.e.,
the relative velocities of the two stars remains the same. In this case, the
condition for stability becomes
∆M
M o
< r
2a o
.
If the orbit is initially circular, the RHS of the above is simply 1/2. If the
initial orbit is elliptical, the RHS of the above is in the range (1 − e o )/2 −
(1 + e o )/2, depending on the precise value of r when the explosion occurs.
In the limit that the initial eccentricity is unity, it is therefore possible
both that very little mass loss could disrupt the orbit, if the stars are near
apastron, or that the binary could remain stable in spite of losing almost
all the binary’s mass, if the stars are near periastron.
If the supernova remnant star receives a kick velocity, which is directed
at an angle of cosθ with respect to ⃗ V o , we find
⃗V = ⃗ V o + ⃗ ∆V , ⃗ V · ⃗ V = V 2 = V 2 o + 2V o ∆V + (∆V ) 2 .
Therefore, the condition for stability becomes
∆M
< r [
1 − ∆V (∆V + 2V ]
o cos θ)
M o 2a o Vc
2 .
In the case that the kick is exactly (mis)aligned with ⃗ V o , we have that
cos θ = +1(−1). Comparing the limiting amounts of mass loss for stability
in the two cases, denoted ∆M + and ∆M − , respectively, we find
∆M + − ∆M −
M o
= − 2r
a o
V o ∆V
V 2 c
< 0.
This means that a kick in the same direction as the orbital motion tends
to destabilize the orbit, while a misaligned kick stabilizes the orbit. In the
case of a circular initial orbit and a misaligned kick, one has
∆M
M o
= 1 2
[
1 + ∆V
V 2 c
]
(2V c − ∆V ) .
If the kick velocity magnitude is comparable to the relative orbital speed in
this case, one has ∆M < M o for stability, i.e., no amount of mass loss can
disrupt the orbit! An oppositely directed kick of this size, however, always
disrupts the binary.